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STRUCTURE OF Ag-105, Ag-107, Ag-109, Ag-110
By Prof. L. Kaliambos (Natural philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favor of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding and the correct nuclear structure by applying the laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Nuclear structure of silver (Ag) with 22 blank positions Naturally occurring silver (Ag) is composed of the two stable isotopes Ag-107 and Ag-109 with Ag-107 being the more abundant (51.839% natural abundance). Twenty-eight radioisotopes have been characterized with the most stable being Ag-105 with a half-life of 41.29 days, Ag-111 with a half-life of 7.45 days, and Ag-112 with a half-life of 3.13 hours. In general Ag with 47 protons (odd number of protons) brakes the high symmetry of Palladium. (See my STRUCTURE OF Pd-102....Pd-110 ). Since the shape of Pd with two squares is elongated the additional p47n47 with the p45n45 of the Pd makes an alpha particle with S=0. Thus for symmetrical arrangements the p40n40 of the square of the Pd is moved to make the symmetrical alpha particle with the p46n46. In the following diagram of Ag you can see the first alpha particle with p45, n45, p47 and n47 and the symmetrical one with the p46, n46, p40 and n40. You can see also the p45, n47, n46 and p40 by using the top view of the third horizontal plane. Note that the two alpha particles contribute to the creation of 8 blank positions able to receive 4(n) +4n. Especially 4(n) give the combinations of p45 and p41, of p47 and p42, of p46 and p43, and of p40 and p44. Whereas the combinations of p45 and p23, of p47 and p29, of p46 and p30, and of p40 and p24. Give the extra 4n which are shown in the diagram near p23, p29, p30 and p24 respectively. In this new arrangement the square of p37n37 and p39n39 with S = -2 gives the same 4 blank positions as those of Pd for receiving 4n of strong bonds. Whereas the symmetrical square of Pd here became the deuteron p38n38 of S = +1 giving 3 blank positions able to receive 2n of strong bonds. As in the case of Pd here the first and the sixth horizontal plane form 4 blank positions for receiving 4(n) of weak bonds, while in the second and the fifth horizontal planes can exist four blank positions for receiving 4{n} with three bonds per neutron. In other worda the number N = 22 of blank positions able to receive extra neutrons is given by The two alpha particles with 4{n) + 4n The horizontal square with 4n The deuteron p39n39 with 2n The first and sixth plane with 4(n) The second and fifth plane with 4{n} That is N = 4(n) +4n + 4n + 2n + 4(n) + 4{n} of opposite spins Or N = 4{n} + 10n + 8(n) = 22 extra neutrons of opposite spins STRUCTURE OF Ag-105, Ag-107 AND Ag -109 WITH S = -1/2 Since in the presence of extra neutrons in the Ag-94 the two alpha particles give S=0 one concludes that the spin S = -1/2 of the unstable Ag-105 is due to the S = -2 of the square, to the S =+1 of the deuteron p38n38 and to the total spin S = +1/2 of the 9 extra neutrons . Therefore, Ag-105 must have 5 extra neutrons of positive spins and 4 extra neutrons of negative spins. That is for Ag -105 we have S = -2 +1 + 5(+1/2) + 4(-1/2) = -1/2 However the instability of Ag-105 is due to this small number of 9 extra neutrons which gives extra np bonds unable to overcome the pp repulsions of long range. Fortunately the stability of the next Ag-107 is due to the 11 extra neutrons . Similarly the stability of Ag- 109 is due to the 13 extra neutrons. In these cases the Ag-107 of 11 extra neutrons must have 6 extra neutrons of positive spins and 5 extra neutrons of negative spins. That is S = -2 +1 +6(+1/2) + 5(-1/2) = -1/2 Also theAg-109 with 13 extra neutrons must have 7 extra neutrons of positive spins and 6 extra neutrons of negative spins. That is S = -2 +1 +7(+1/2) + 6(-1/2) = -1/2 Since N = 4{n} + 10n + 8(n) of opposite spins one concludes that that the 7 extra neutrons of positive spins contain 2{n} and 5n , while the 6 extra neutrons contain 2{n} and 4n. That is the structure has not extra (n) of weak bonds. STRUCTURE OF THE UNSTABLE Ag-110 WITH S = +1 In the unstable structure of Ag-110 with S=+1 the additional p47n47 fills the blank position behind the p40n40. This situation reduces the number of extra neutrons n and the Ag-110 receives 14 extra neutrons of opposite spins including several (n) of weak bonds. ' ' ' DIAGRAM OF SILVER -94 WITH S = -1 HAVING 22 BLANK POSITIONS' Here you see the additional p47n47 and also the p40n40 which changes the spin from S = +1 to S =0 giving S = -1 for making symmetrical alpha particle of opposite spins . Whereas the p41, n41, p42, n42, p43, n43 p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Note that the 4 extra neutrons (n) of the first and the sixth plane along with the 3 extra neutrons n near the p37, p38 and p39 are not shown. You can see only the 3 extra neutrons n existing under the p21 and p22 and over the p31. ' ' ' ' n........p38..........n38 p38n38 with n ' ' n31………p12.........n12.......p32 ' p31........n11.........p11…… n32 Sixth H. plane' ' n......... p29.........n10........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n40' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p40' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24...........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n' ........p37.. ....n37 ' ' n39.......p39........n ' H. square with n' ' ' ' ' ' TOP VIEW OF THE SQUARE OF p37n37 AND n39p39 ' THE TWO EXTRA NEUTRONS n WITH STRONG VERTICAL BONDS UNDER THE p21 AND p22 ARE ALSO SHOWN IN THE DIAGRAM OF Rh , WHILE THE TWO EXTRA NEUTRONS nWITH STRONG VERTICAL BONDS UNDER THE p34 AND p33 ARE NOT SHOWN IN THE DIAGRAM ' n' ' n........P37.........n37 ' ' n39........p39.........n ' ' n' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ' ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' TOP VIEW OF THE THIRD HORIZONTAL PLANE OF POSITIVE SPINS WITH THE 4 NUCLEONS LIKE p41, n43, n42 AND p44 WHICH MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED. HERE THE p45 AND n47 ALONG WITH THE n46 AND p40 MAKE THE SYMMETRICAL SQUARES OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n15p15 AND p16n16. ' ' n42........p16......n16......p44 ' n47........p25........n6........p6........n26.........p40' ' p45........n25........p5........n5........p26........ n40' ' p41.......n15.......p15.......n43' Category:Fundamental physics concepts